18.6 Convergence in Measure
Definition 18.6.1 (Convergence in Measure). Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$, then $f_{n} \to f$ in measure if for every $\eps > 0$,
Definition 18.6.2 (Cauchy in Measure). Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ be Borel measurable functions from $X$ to $Y$, then $\seq{f_n}$ is Cauchy in measure if for every $\eps > 0$,
Lemma 18.6.3. Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$ such that $f_{n} \to f$ almost everywhere, then $f_{n} \to f$ in measure.
Proof. Let $\eps > 0$, then for almost every $x \in X$, there exists $N \in \natp$ such that $d(f_{n}(x), f(x)) < \eps$ for all $n \ge N$, so
By continuity from above (Proposition 15.1.5),
Proposition 18.6.4. Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a complete metric space, and $\seq{f_n}$ be Borel measurable functions from $X \to Y$, then:
There exists a Borel measurable function $f: X \to Y$ such that $f_{n} \to f$ in measure.
There exists a subsequence $\seq{n_k}$ such that $f_{n_k}\to f$ almost everywhere.
For any Borel measurable function $g: X \to Y$ such that $f_{n} \to g$ in measure, $f = g$ almost everywhere.
Proof. (2): Since $\seq{f_n}$ is Cauchy in measure, there exists a subsequence $\seq{n_k}$ such that for each $k \in \natp$, $\mu(\bracsn{d(f_{n_k}, f_{n_{k+1}}) > 2^{-k}}) \le 2^{-k}$.
In this case, for any $K \in \natp$ and $j \ge k \ge K$,
By monotonicity and subadditivity (Proposition 15.1.5),
so
By the First Borel-Cantelli Lemma,
Thus, for almost every $x \in X$, there exists $K \in \natp$ such that $d(f_{n_j}(x), f_{n_k}(x)) < 2^{-K+1}$ for all $j, k \ge K$. Therefore $\seq{f_n(x)}$ is Cauchy for almost every $x$, and converges to a Borel measurable function $f: X \to Y$.
(1): Let $\eps, \delta > 0$, then there exists $N \in \natp$ such that $\mu(\bracs{d(f_m, f_n) > \delta/2}) < \eps/2$ for all $m, n \ge N$. Let $k \in \natp$ such that $n_{k} \ge N$ and $\mu(\bracs{d(f, f_{n_k}) > \delta/2}) < \eps/2$, then for any $m \ge N$,
(3): By (2), there exists a subsequence $\seq{n_k}$ such that $f_{n_k}\to f$ and $f_{n_k}\to g$ almost everywhere, so $f = g$ almost everywhere.$\square$